結果

問題 No.872 All Tree Path
ユーザー HaarHaar
提出日時 2019-08-30 22:12:11
言語 C++17
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 214 ms / 3,000 ms
コード長 3,217 bytes
コンパイル時間 2,125 ms
コンパイル使用メモリ 204,372 KB
実行使用メモリ 27,280 KB
最終ジャッジ日時 2023-08-14 05:30:47
合計ジャッジ時間 5,061 ms
ジャッジサーバーID
(参考情報)
judge11 / judge12
このコードへのチャレンジ
(要ログイン)

テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 199 ms
19,852 KB
testcase_01 AC 214 ms
19,908 KB
testcase_02 AC 209 ms
20,112 KB
testcase_03 AC 88 ms
27,280 KB
testcase_04 AC 1 ms
4,376 KB
testcase_05 AC 210 ms
19,944 KB
testcase_06 AC 210 ms
19,888 KB
testcase_07 AC 209 ms
19,908 KB
testcase_08 AC 12 ms
4,792 KB
testcase_09 AC 13 ms
4,856 KB
testcase_10 AC 12 ms
4,816 KB
testcase_11 AC 12 ms
4,768 KB
testcase_12 AC 13 ms
4,860 KB
testcase_13 AC 2 ms
4,380 KB
testcase_14 AC 1 ms
4,380 KB
testcase_15 AC 2 ms
4,376 KB
testcase_16 AC 2 ms
4,376 KB
testcase_17 AC 2 ms
4,380 KB
testcase_18 AC 1 ms
4,380 KB
testcase_19 AC 2 ms
4,376 KB
権限があれば一括ダウンロードができます

ソースコード

diff #

#include <bits/stdc++.h>
#define LLI long long int
#define FOR(v, a, b) for(LLI v = (a); v < (b); ++v)
#define FORE(v, a, b) for(LLI v = (a); v <= (b); ++v)
#define REP(v, n) FOR(v, 0, n)
#define REPE(v, n) FORE(v, 0, n)
#define REV(v, a, b) for(LLI v = (a); v >= (b); --v)
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define ITR(it, c) for(auto it = (c).begin(); it != (c).end(); ++it)
#define RITR(it, c) for(auto it = (c).rbegin(); it != (c).rend(); ++it)
#define EXIST(c,x) ((c).find(x) != (c).end())
#define fst first
#define snd second
#define popcount __builtin_popcount
#define UNIQ(v) (v).erase(unique(ALL(v)), (v).end())
#define bit(i) (1LL<<(i))

#ifdef DEBUG
#include <misc/C++/Debug.cpp>
#else
#define dump(...) ((void)0)
#endif

#define gcd __gcd

using namespace std;
template <class T> constexpr T lcm(T m, T n){return m/gcd(m,n)*n;}

template <typename I> void join(ostream &ost, I s, I t, string d=" "){for(auto i=s; i!=t; ++i){if(i!=s)ost<<d; ost<<*i;}ost<<endl;}
template <typename T> istream& operator>>(istream &is, vector<T> &v){for(auto &a : v) is >> a; return is;}

template <typename T, typename U> bool chmin(T &a, const U &b){return (a>b ? a=b, true : false);}
template <typename T, typename U> bool chmax(T &a, const U &b){return (a<b ? a=b, true : false);}
template <typename T, size_t N, typename U> void fill_array(T (&a)[N], const U &v){fill((U*)a, (U*)(a+N), v);}

struct Init{
  Init(){
    cin.tie(0);
    ios::sync_with_stdio(false);
  }
}init;

template <typename Cost = int> class Edge{
public:
  int from,to;
  Cost cost;
  Edge() {}
  Edge(int to, Cost cost): to(to), cost(cost){}
  Edge(int from, int to, Cost cost): from(from), to(to), cost(cost){}

  Edge rev() const {return Edge(to,from,cost);}
  
  friend ostream& operator<<(ostream &os, const Edge &e){
    os << "(FROM: " << e.from << "," << "TO: " << e.to << "," << "COST: " << e.cost << ")";
    return os;
  }
};

template <typename T> using Graph = vector<vector<Edge<T>>>;
template <typename T> using Tree = vector<vector<Edge<T>>>;

template <typename C, typename T> void add_edge(C &g, int from, int to, T w){
  g[from].push_back(Edge<T>(from, to, w));  
}

template <typename C, typename T> void add_undirected(C &g, int a, int b, T w){
  g[a].push_back(Edge<T>(a, b, w));
  g[b].push_back(Edge<T>(b, a, w));
}

Tree<int> tree;
int N;

int sub[202000];
LLI dp1[202000];
LLI dp2[202000];


void dfs(int cur, int par){
  dp1[cur] = 0;
  sub[cur] = 1;
  
  for(auto &e : tree[cur]){
    if(e.to == par) continue;

    dfs(e.to, cur);

    sub[cur] += sub[e.to];
    dp1[cur] += dp1[e.to] + sub[e.to] * e.cost;
  }  
}

void dfs2(int cur, int par, LLI cost){

  if(par != -1){
    dp2[cur] += dp2[par];
    
    dp2[cur] += dp1[par] - (dp1[cur] + sub[cur] * cost);
  }

  dp2[cur] += (N-sub[cur]) * cost;

  for(auto &e : tree[cur]){
    if(e.to == par) continue;
    dfs2(e.to, cur, e.cost);
  }
}



int main(){
  cin >> N;
  tree = Tree<int>(N);
  REP(i,N-1){
    int u,v,w; cin >> u >> v >> w;
    --u, --v;
    add_undirected(tree, u, v, w);
  }
  
  dfs(0,-1);
  dfs2(0,-1,0);
  
  LLI ans = 0;


  REP(i,N) ans += dp1[i] + dp2[i];


  cout << ans << endl;

  return 0;
}
0